# Find the circumcenter of a triangle with the given vertices. #1) H(5,0), J(0,3), K(0,0) #2) L(0,0), M(-2, 0), N(0, -4)

*print*Print*list*Cite

1) H (5,0) J(0,3) K(0,0)

m1 = slope HJ = ( 3-0)/(0-5) = -3/5

M1 = midpoint HJ = ( 5/2, 3/2)

Then the perpendicular bisector is:

y- 3/2 = 5/3( X-5/2)

Y- 3/2 = (5/3)X - 25/6

Y= (5/3)X - 25/6 + 3/2

Y= (5/3)x - 8/3................(1)

m2= slope HK = 0 ==> y= 5 ==> x

M2 = midpoint HK = ( 5/2, 0)

==> The equation of the line is

x = 5/2.............(2)

Now we will substitute in (1):

==> y= (5/3)x - 8/3

==> y= (5/3) *5/2 - 8/3

= (25/6 - 8/3

= 3/2

**Then the cir-cum-center is ( 5/2 , 3/2)**

To verify we will find the equation of the perpendicular bisector of JK

m3 = slope JK = ( 3/0)

==> the equation is y= 3/2

1)

We see that if we plot the coordinates on the graph, the triangle HKJ is a right angled triangle , with right angle at K(0,0) and HJ as hypotenuse. Therefore the mid point of the hypotenuse HJ is the circumcentre . The coordinates of H and J are : H (5,0) and J(0,3).

Therefore the mid point of HJ is given by :

Mx = (Hx+Jx)/2 = ((5+0)/2 = 2.5.

My = (Hy + Jy)/2 = (0+3)/2 = 1.5.

Therefore the circumcentre of the triangle HKJ is (2.5 , 1.5).

2)

The triangle MLN is a right angled triangle with right angle at L(0,0). The hypotenuse of the triangle is MN with M at (-2,0) and N at (0,4).

So the mid point of the hypotenuse MN is the circumcentre.

The mid point C of MN has the coordinates given by:

Cx = (Mx+Nx)/2 = ((-2+0)/2 = -1.

Cy = (My+Ny)/2 = ((0-4)/2 = =-2.

Therefore the circumcentre of the triangle MLN is (-1,-2).

`Delta` HJK is a Right angled triangle

So, circumcentre is the midpoint of hypotenuse(HJ)

the circumcentre is (5/2 , 3/2) (or) (2.5,1.5).

`Delta` LMN is also a right angled triangle

So, circumcentre is the midpoint of hypotenuse(MN)

the circumcetre is (-1,-2)